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Cloud [144]
4 years ago
6

The chart shows the bids provided by four engineers to test a prototype.

Engineering
1 answer:
klasskru [66]4 years ago
8 0

Answer:

D

Explanation:

To know which is most or least cost-effective, it's not enough to look at only the per day rate, or only the time to complete.  You have to multiply them to get the total cost of the project.

\left[\begin{array}{ccccc}&Cost\ per\ day\ (\$)&Time\ to\ complete\ (days)&Total\ cost\ (\$)\\Zoe&500&8&4000\\Greg&650&10&6500\\Orion&400&12&4800\\Jin&700&5&3500\end{array}\right]

As you can see, Greg is the least cost-effective because he charges the most for the project.

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What are the four categories of engineering materials used in manufacturing?
alexgriva [62]

Answer:

metals, composite, ceramics and polymers.

Explanation:

The four categories of engineering materials used in manufacturing are metals, composite, ceramics and polymers.

i) Metals: Metals are solids made up of atoms held by matrix of electrons. They are good conductors of heat and electricity, ductile and strong.

ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.

iii) Ceramics: They are inorganic, non-metallic crystalline compounds with high hardness and strength as well as poor conductors of electricity and heat.

iv) Polymers: They  have low weight and are poor conductors of electricity and heat

8 0
3 years ago
A good visual lead is .... seconds from the front of the vehicle, focusing in the center of the path of travel. Searching 20 to
Tomtit [17]

A good visual lead is 20-30 seconds from the front of the vehicle, focusing in the center of the path of travel. Searching 20 to 30 seconds ahead, gives you time to assess within the next ..12-15 seconds, actions you may need to take to control an approaching risk.

Discussion:

Keeping eye focus centered in a path of travel at an interval of 20 to 30 seconds away from the vehicle is critical to gaining enough info. as possible in the driving scene. Good targeting sets up good sight lines for referencing and good peripheral fields for observing changes.

  • It is important to look ahead 12-15 seconds into your target area as one drives. Compromise. space by giving as much space to the greater of two hazards.

Read more on driving visual leads:

brainly.com/question/7067386

3 0
3 years ago
Thermal conductivity of AISI 316 Stainless Steel at 90ºC is 14.54 W/m K. Convert this value to IP system.
abruzzese [7]

Answer:

the value of conductivity in IP is 8.406\dfrac{Btu}{ft.hr.F}

Explanation:

Given that

Thermal conductivity K=14.54 W/m.K

This above given conductivity is in SI unit.

     SI unit                                           IP unit              Conversion factor

    m                                                      ft                      0.3048

   W                                                       Btu/hr               0.293          

 

The unit of conductivity in IP is Btu./ft.hr.F.

Now convert into IP divided by 1.73 factor.

0.57\dfrac{Btu}{ft.hr.F}=1 \dfrac{W}{m.K}

So

0.57\times 14.54\dfrac{Btu}{ft.hr.F}=14.54 \dfrac{W}{m.K}

8.406\dfrac{Btu}{ft.hr.F}=14.54 \dfrac{W}{m.K}

So the value of conductivity in IP is 8.406\dfrac{Btu}{ft.hr.F}

 

7 0
3 years ago
A small lake with volume of 160,000 m^3 receives agricultural drainage waters that contain 150 mg / L total dissolved solids (TD
Stels [109]

Answer:

Explanation:

Given that : -

The desirable limit is 500 mg / l , but

allowable upto 2000 mg / l.

The take volume is V = 160.000 m3

V = 160 , 000 x 103 l

The crainage gives 150 mg / l and lake has initialy 100 mg / l

Code of tpr frpm drawn = 150 x 60, 000 x 1000

Ci = 9000 kg / gr

Cl = 100 x 160,000 x 1000

Cl = 16, 000 kg

Since allowable limit = 2000 mg / l

Cn = ( 2000 x 160, 00 x 1000 )

= 320, 000 kg

so, each year the rate increases, by 9000 kg / yr

Read level = ( 320, 000 - 16,000 )

Li = 304, 000 kg

Tr=<u>304,000</u>

      900

=33.77

5 0
3 years ago
Read 2 more answers
plane wall, 7.5 cm thick, generates heat internally at the rate of 105W/m3. One side of the wall is insulated, and the other sid
valentinak56 [21]

Answer:

T=120.04°C

Explanation:

Given that

L= 7.5 cm

q = 105 W/m³

T∞=120°C

h=750 W/m²K

K=20 W/mK

Here given that one side of the wall is insulated that is why the maximum temperature will be at the insulated surface.

The total heat transfer from the wall

Q= q A L

Q= 150 x 0.075 A

Q=7.875 A W

A=Area of wall

Now the total thermal resistance R

R=\dfrac{L}{KA}+\dfrac{1}{hA}

R=\dfrac{0.075}{20A}+\dfrac{1}{750A}

R=\dfrac{0.00508}{A}

We also know that

Q=\dfrac{\Delta T}{R}

Temperature at insulated side = T

7.875 A=\dfrac{T-120}{\dfrac{0.00508}{A}}

7.875 =\dfrac{T-120}{0.00508}

T=120.04°C

5 0
4 years ago
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